What is the tangent of 1/3 x^3+7/2 x^2-3x+5 ?

The tangent of 1/3 x^3+7/2 x^2-3x+5 is y=(a_{0}^2+7a_{0}-3)x-2/3 a_{0}^3-7/2 a_{0}^2+5

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tangent of 1/3 x^3+7/2 x^2-3x+5

tangent of

\frac{1}{3} x^{3} + \frac{7}{2} x^{2} - 3 x + 5

y = (a_{0}^{2} + 7 a_{0} - 3) x - \frac{2}{3} a_{0}^{3} - \frac{7}{2} a_{0}^{2} + 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{3} a_{0}^{3} + \frac{7}{2} a_{0}^{2} - 3 a_{0} + 5)

Find the slope of

y = \frac{1}{3} x^{3} + \frac{7}{2} x^{2} - 3 x + 5 : \frac{d y}{d x} = x^{2} + 7 x - 3

Compute the slope of

y = \frac{1}{3} x^{3} + \frac{7}{2} x^{2} - 3 x + 5

x = a_{0} : m = a_{0}^{2} + 7 a_{0} - 3

Find the line with slope m=

a_{0}^{2} + 7 a_{0} - 3

and passing through

(a_{0}, \frac{1}{3} a_{0}^{3} + \frac{7}{2} a_{0}^{2} - 3 a_{0} + 5) : y = (a_{0}^{2} + 7 a_{0} - 3) x - \frac{2}{3} a_{0}^{3} - \frac{7}{2} a_{0}^{2} + 5

y = (a_{0}^{2} + 7 a_{0} - 3) x - \frac{2}{3} a_{0}^{3} - \frac{7}{2} a_{0}^{2} + 5

f (t) = tan (t)

x \to 1 lim (e^{x^{2} - x})

\int sin^{12} (x) d x

\frac{d}{d x} (x - 5 sin (x))

\frac{d ^{2}}{d x ^{2}} (x + 10)

What is the tangent of 1/3 x^3+7/2 x^2-3x+5 ?
The tangent of 1/3 x^3+7/2 x^2-3x+5 is y=(a_{0}^2+7a_{0}-3)x-2/3 a_{0}^3-7/2 a_{0}^2+5